Extra Credit 16

Q.19.5A

Determine whether a forward reaction is spontaneous or non-spontaneous:

1. \(\mathrm{Ag^{2+}(aq) + Cr^{2+}(aq) \rightarrow Ag^+ (aq)}\)
2. \(\mathrm{Sn^{4+}(aq) + 2I^-(aq) \rightarrow Sn^{2+} (aq) + I_2 (aq)}\)

S19.5A

1. \(\mathrm{E^\circ_{cell}= E^\circ_{cathode} - E^\circ_{anode}}\)

\(\mathrm{E^\circ_{cell}= (E^\circ_{Ag^{2+}/Ag^-}) - (E^\circ_{Cr^{3+}/Cr^{2+}})}\)
\(\mathrm{E^\circ_{cell}= 1.98\,V - (-0.424\,V)}\)
\(\mathrm{E^\circ_{cell}= +2.404\,V}\)
The forward reaction yields a spontaneous reaction.

2. \(\mathrm{Sn^{4+}(aq)\,reduced,\: 2I^-(aq)\,oxidized}\)

\(\mathrm{E^\circ_{cell} = (+0.154\,V) - (+0.535\,V)}\)
\(\mathrm{E^\circ_{cell} = -0.381\,V}\)
The forward reaction yields a nonspontaneous reaction.

Q19.45A

Solve for \(\mathrm{E_{cell}}\) of the following voltaic cell

\(\mathrm{Cu(s) | Cu^{2+} (0.01\:M) || Cu^{2+} (0.1\:M) | Cu(s)}\)

S19.45A

We first solve for \(\mathrm{E^\circ_{cell}}\)

Oxidation: \(\mathrm{Cu(s) \rightarrow Cu^{2+} + 2e^- \quad E^\circ = -0.340}\)

Oxidation is when the electrons are loss in the process of the reaction.

Reduction: \(\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \quad E^\circ= +0.340}\)

Reduction is when the electrons are gained in the process of the reaction.

\(\mathrm{E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}}\)

\(\mathrm{0.340-0.340 = 0}\)

Then, using the Nernst Equation to solve for \(\mathrm{E_{cell}}\), we derive

\(\begin{align}
\mathrm{E_{cell}} & = \mathrm{E^\circ_{cell} - \dfrac{0.0257}{2}\ln Q} \\
& = \mathrm{0 - \dfrac{0.0257}{2}\ln\dfrac{0.01\:M}{0.1\:M}} \\
\mathrm{E_{cell}} & = \mathrm{0.0296\:V}
\end{align}\)

Q21.3A

Sketch the geometric isomers of \(\ce{[CoBr2(en)(CO)2]-}\).

Q24.5A

In the reaction \(\mathrm{A \rightarrow products}\), 4.50 minutes after the reaction is started, \(\mathrm{[A]=0.587\,M}\). The rate of reaction at this point is \(\mathrm{rate = -\dfrac{\Delta[A]}{\Delta t} = 2.1 \times 10^{-2}\, M\, min^{-1}}\). Assume that this rate remains constant for a short period of time.

1. What is \(\mathrm{[A]}\) 6.00 minutes after the reaction is started?
2. At what time after the reaction is started with \(\mathrm{[A] = 0.56\,M}\)?

Q24.47A

For the reversible reaction \(\ce{A + B \leftrightarrow A + B}\) the enthalpy change of the forward reaction is +11 kj/mol. The activation energy of the forward reaction is 74 kj/mol.

What is the activation energy for the reverse reaction?

S24.47A

To go from the products to reactants, you would have to add the 11 KJ of energy just to get back to the original energy of the reactants. But you also have to add in the extra activation energy to get back where you started from. Thinking about the conservation of energy is helpful here. If the reactants have 11KJ less energy than the products, and you had to add another 74KJ of energy just to get back to the reactants, then the total change must be 63KJ.

Q25.25C

Suppose a sample with \(\ce{^{170}_{69}Tm}\) has an activity 100 times the detectable limit. How long until the sample's radioactivity is no longer detectable?

S25.25C

\(\mathrm{\ce{^{170}_{69}Tm}\: half\: life = 128.6\: days}\)

\(\mathrm{\lambda = \dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{128.6\,d}=0.00539\,d^{-1}}\)

\(\mathrm{\ln\dfrac{1}{100} = -0.00539\,d^{-1}(t)}\)

\(\mathrm{t=854\: days}\)

Q18.3

Classify each of the following substances as an oxidizing agent, reducing agent or both. List the oxidizing agents in order of decreasing strength; list the reducing agents in order of decreasing strength (use SRP Table):

\(\ce{Ni(s)}\), \(\ce{H+ (aq)}\), \(\ce{Au(s)}\), \(\ce{Cl2(g)}\), \(\ce{Sn^2+ (aq)}\), \(\ce{Mg(s)}\), \(\ce{Fe^2+ (aq)}\)

Q21.3.9

When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium?